For the reaction ${Ni}^{2 +} + 4 {NH}_{3} ⟶ {[Ni {({NH}_{3})}_{4}]}^{2 +}$ ; at equilibrium, if the solution contains $1.6 \times 10^{- 4 %}$ of nickel ion in the free state, and the concentration of NH₃ at equilibrium is 0.5 M. Then the instability constant of the complex will be approximately equal

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$1.5 \times 10^{- 16}$

$1.0 \times 10^{7}$

$1.0 \times 10^{- 5}$

$1.5 \times 10^{- 17}$

answer is C.

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Detailed Solution

For given reaction,

$\begin{array}{l} {Ni}^{2 +} + 4 {NH}_{3} \Leftrightarrow {[Ni {({NH}_{3})}_{4}]}^{2 +} \\ K = \frac{{[Ni {({NH}_{3})}_{4}]}^{2 +}}{[{Ni}^{2 +} {[{NH}_{3}]}^{4}]} \end{array}$

$\begin{array}{l} \frac{[{Ni}^{2 +}]}{[{Ni}^{2 +}] + {[Ni {(N_{3})}_{4}]}^{2 +}} = 1 .6 \times 10^{- 6} \\ K = \frac{10^{6}}{1 .6 \times (0 .5)^{4}} \\ K = 10^{7} \end{array}$